Implement the type version of Array.indexOf, indexOf<T, U> takes an Array T, any U and returns the index of the first U in Array T.
type Res = IndexOf<[1, 2, 3], 2>; // expected to be 1
type Res1 = IndexOf<[2,6, 3,8,4,1,7, 3,9], 3>; // expected to be 2
type Res2 = IndexOf<[0, 0, 0], 2>; // expected to be -1
import type { Equal, Expect } from '@type-challenges/utils'
type cases = [
Expect<Equal<IndexOf<[1, 2, 3], 2>, 1>>,
Expect<Equal<IndexOf<[2, 6, 3, 8, 4, 1, 7, 3, 9], 3>, 2>>,
Expect<Equal<IndexOf<[0, 0, 0], 2>, -1>>,
Expect<Equal<IndexOf<[string, 1, number, 'a'], number>, 2>>,
Expect<Equal<IndexOf<[string, 1, number, 'a', any], any>, 4>>,
Expect<Equal<IndexOf<[string, 'a'], 'a'>, 1>>,
Expect<Equal<IndexOf<[any, 1], 1>, 1>>,
]